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3.2: Basic Concepts

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Basic Concepts

A $$\textcolor{blue}{\mbox{differential equation}}$$ is an equation that contains one or more derivatives of an unknown function. The $$\textcolor{blue}{\mbox{order}}$$ of a differential equation is the order of the highest derivative that it contains. A differential equation is an $$\textcolor{blue}{\mbox{ordinary differential equation}}$$ if it involves an unknown function of only one variable, or a $$\textcolor{blue}{\mbox{partial differential equation}}$$ if it involves partial derivatives of a function of more than one variable. For now we'll consider only ordinary differential equations, and we'll just call them $$\textcolor{blue}{\mbox{differential equations}}$$.

Throughout this text, all variables and constants are real unless it's stated otherwise. We'll usually use $$x$$ for the independent variable unless the independent variable is time; then we'll use $$t$$.

The simplest differential equations are first order equations of the form

\begin{eqnarray*}
{dy\over dx}=f(x) \mbox{ or, equivalently, }y'=f(x),
\end{eqnarray*}

where $$f$$ is a known function of $$x$$. We already know from calculus how to find functions that satisfy this kind of equation. For example, if

\begin{eqnarray*}
y'=x^3,
\end{eqnarray*}

then

\begin{eqnarray*}
y=\int x^3\, dx={x^4\over4}+c,
\end{eqnarray*}

where $$c$$ is an arbitrary constant. If $$n>1$$ we can find functions $$y$$ that satisfy equations of the form

$$y^{(n)}=f(x)$$

by repeated integration. Again, this is a calculus problem.

Except for illustrative purposes in this section, there's no need to consider differential equations like equation $$(1)$$. We'll usually consider differential equations that can be written as

$$y^{(n)}=f(x,y,y', \dots,y^{(n-1)}),$$

where at least one of the functions $$y$$, $$y'$$, $$\dots$$, $$y^{(n-1)}$$ actually appears on the right. Here are some examples:

\begin{eqnarray*}
\displaystyle {dy\over dx}-x^2&=&0&\mbox{ (first order)}, \\
\displaystyle{dy\over dx}+2xy^2&=&-2&\mbox{ (first order)}, \\
\displaystyle{d^2y\over dx^2}+2\displaystyle{dy\over dx}+y&=&2x&\mbox{ (second order)}, \\
xy'''+y^2&=&\sin x &\mbox{ (third order)}, \\
y^{(n)}+xy'+3y&=&x&\mbox{ ($n$-th order)}.
\end{eqnarray*}

Although none of these equations is written as in equation $$(2)$$, all of them $$\textcolor{blue}{\mbox{can}}$$ be written in this form:

\begin{eqnarray*}
y'&=&x^2, \\
y'&=&-2-2xy^2, \\
y''&=&2x-2y'-y, \\
y'''&=&\displaystyle{\sin x-y^2\over x}, \\
 y^{(n)}&=&x-xy'-3y.
\end{eqnarray*}

Solutions of Differential Equations

A $$\textcolor{blue}{\mbox{solution}}$$ of a differential equation is a function that satisfies the differential equation on some open interval; thus, $$y$$ is a solution of equation $$(2)$$ if $$y$$ is $$n$$ times differentiable and

\begin{eqnarray*}
y^{(n)}(x)=f(x,y(x),y'(x), \dots,y^{(n-1)}(x))
\end{eqnarray*}

for all $$x$$ in some open interval $$(a,b)$$. In this case, we also say that $$y$$ $$\textcolor{blue}{\mbox{ is a solution of equation \( (2)$$ on}} \) $$(a,b)$$. Functions that satisfy a differential equation at isolated points are not interesting. For example, $$y=x^2$$ satisfies

\begin{eqnarray*}
xy'+x^2=3x
\end{eqnarray*}

if and only if $$x=0$$ or $$x=1$$, but it's not a solution of this differential equation because it does not satisfy the equation on an open interval.

The graph of a solution of a differential equation is a $$\textcolor{blue}{\mbox{solution curve}}$$. More generally, a curve $$C$$ is said to be an $$\textcolor{blue}{\mbox{integral curve}}$$ of a differential equation if every function $$y=y(x)$$ whose graph is a segment of $$C$$ is a solution of the differential equation. Thus, any solution curve of a differential equation is an integral curve, but an integral curve need not be a solution curve.

Example $$\PageIndex{1}$$:

If $$a$$ is any positive constant, the circle

$$x^2+y^2=a^2$$

is an integral curve of

$$y'=-{x\over y}.$$

To see this, note that the only functions whose graphs are segments of equation $$(3)$$ are

\begin{eqnarray*}
y_1=\sqrt{a^2-x^2}\mbox{ and } y_2=-\sqrt{a^2-x^2}.
\end{eqnarray*}

We leave it to you to verify that these functions both satisfy equation $$(4)$$ on the open interval $$(-a,a)$$. However, equation $$(3)$$ is not a solution curve of equation $$(4)$$ since it's not the graph of a function.

Example $$\PageIndex{2}$$:

Verify that

$$y={x^2\over3}+{1\over x}$$

is a solution of

$$xy'+y=x^2$$

on $$(0,\infty)$$ and on $$(-\infty,0)$$.

Substituting equation $$(5)$$ and

\begin{eqnarray*}
y'={2x\over3} - {1\over x^2}
\end{eqnarray*}

into equation $$(6)$$ yields

\begin{eqnarray*}
xy'(x)+y(x)=x \left({2x\over3} - {1\over x^2}\right)+\left({x^2\over3}+{1\over x}\right)=x^2
\end{eqnarray*}

for all $$x\ne0$$. Therefore $$y$$ is a solution of equation $$(6)$$ on $$(-\infty,0)$$ and $$(0,\infty)$$. However, $$y$$ isn't a solution of the differential equation on any open interval that contains $$x=0$$, since $$y$$ is not defined at $$x=0$$.

Figure $$1.2.1$$ shows the graph of equation $$(5)$$. The part of the graph of equaiton $$(5)$$ on $$(0,\infty)$$ is a solution curve of equation $$(6)$$, as is the part of the graph on $$(-\infty,0)$$.

Example $$\PageIndex{3}$$:

Show that if $$c_1$$ and $$c_2$$ are constants then

$$y=(c_1+c_2x)e^{-x}+2x-4$$

is a solution of

$$y''+2y'+y=2x$$

on $$(-\infty,\infty)$$. $$\textcolor{blue}{\mbox{ Figure \(1.2.1$$: $$y=\displaystyle{\frac{x^{2}}{3}+\frac{1}{x}}$$ }} \)

Differentiating equation $$(7)$$ twice yields

\begin{eqnarray*}
y'=-(c_1+c_2x)e^{-x}+c_2e^{-x}+2
\end{eqnarray*}

and

\begin{eqnarray*}
y''=(c_1+c_2x)e^{-x}-2c_2e^{-x},
\end{eqnarray*}

so

\begin{eqnarray*}
y''+2y'+y&=&(c_1+c_2x)e^{-x}-2c_2e^{-x} +2\left[-(c_1+c_2x)e^{-x}+c_2e^{-x}+2\right] +(c_1+c_2x)e^{-x}+2x-4\\
&=&(1-2+1)(c_1+c_2x)e^{-x}+(-2+2)c_2e^{-x}+4+2x-4 \\
&=&2x
\end{eqnarray*}

for all values of $$x$$. Therefore $$y$$ is a solution of equation $$(8)$$ on $$(-\infty,\infty)$$.

Example $$\PageIndex{4}$$:

Find all solutions of

$$y^{(n)}=e^{2x}.$$

Integrating equation $$(9)$$ yields

\begin{eqnarray*}
y^{(n-1)}={e^{2x}\over2}+k_1,
\end{eqnarray*}

where $$k_1$$ is a constant. If $$n\ge2$$, integrating again yields

\begin{eqnarray*}
y^{(n-2)}={e^{2x}\over4}+k_1x+k_2.
\end{eqnarray*}

If $$n\ge3$$, repeatedly integrating yields

$$y={e^{2x}\over2^n}+k_1{x^{n-1}\over (n-1)!}+k_2{x^{n-2}\over (n-2)!}+\cdots+k_n,$$

where $$k_1$$, $$k_2$$, $$\dots$$, $$k_n$$ are constants. This shows that every solution of equation $$(9)$$ has the form equation $$(10)$$ for some choice of the constants $$k_1$$, $$k_2$$, $$\dots$$, $$k_n$$. On the other hand, differentiating equation $$(10)$$ $$n$$ times shows that if $$k_1$$, $$k_2$$, $$\dots$$, $$k_n$$ are arbitrary constants, then the function $$y$$ in equation $$(10)$$ satisfies equation $$(9)$$.

Since the constants $$k_1$$, $$k_2$$, $$\dots$$, $$k_n$$ in equation $$(10)$$ are arbitrary, so are the constants

\begin{eqnarray*}
{k_1\over (n-1)!},\, {k_2\over(n-2)!},\, \cdots, \, k_n.
\end{eqnarray*}

Therefore Example $$4$$ actually shows that all solutions of equation $$(9)$$ can be written as

\begin{eqnarray*}
y={e^{2x}\over2^n}+c_1+c_2x+\cdots+c_nx^{n-1},
\end{eqnarray*}

where we renamed the arbitrary constants in equation $$(10)$$ to obtain a simpler formula. As a general rule, arbitrary constants appearing in solutions of differential equations should be simplified if possible. You'll see examples of this throughout the text.

Initial Value Problems

In Example $$4$$ we saw that the differential equation $$y^{(n)}=e^{2x}$$ has an infinite family of solutions that depend upon the $$n$$ arbitrary constants $$c_1$$, $$c_2$$, $$\dots$$, $$c_n$$. In the absence of additional conditions, there's no reason to prefer one solution of a differential equation over another. However, we'll often be interested in finding a solution of a differential equation that satisfies one or more specific conditions. The next example illustrates this.

Example $$\PageIndex{5}$$:

Find a solution of

\begin{eqnarray*}
y'=x^3
\end{eqnarray*}

such that $$y(1)=2$$.

At the beginning of this section we saw that the solutions of $$y'=x^3$$ are

\begin{eqnarray*}
y={x^4\over4}+c.
\end{eqnarray*}

To determine a value of $$c$$ such that $$y(1)=2$$, we set $$x=1$$ and $$y=2$$ here to obtain

\begin{eqnarray*}
\end{eqnarray*}

Therefore the required solution is

\begin{eqnarray*}
y={x^4+7\over4}.
\end{eqnarray*}

Figure $$1.2.2$$ shows the graph of this solution. Note that imposing the condition $$y(1)=2$$ is equivalent to requiring the graph of $$y$$ to pass through the point $$(1,2)$$.

We can rewrite the problem considered in Example $$5$$ more briefly as

\begin{eqnarray*}
\end{eqnarray*}

We call this an $$\textcolor{blue}{\mbox{value problem}}$$. The requirement $$y(1)=2$$ is an $$\textcolor{blue}{\mbox{initial condition}}$$. Initial value problems can also be posed for higher order differential equations. For example,

$$y'' - 2y'+3y=e^x, \quad y(0)=1, \quad y'(0)=2$$

is an initial value problem for a second order differential equation where $$y$$ and $$y'$$ are required to have specified values at $$x=0$$. In general, an initial value problem for an $$n$$-th order differential equation requires $$y$$ and its first $$n-1$$ derivatives to have specified values at some point $$x_0$$. These requirements are the $$\textcolor{blue}{\mbox{initial conditions}}$$. $$\textcolor{blue}{\mbox{ Figure \(1.2.2$$: $$y=\displaystyle{\frac{x^{2}+7}{4}}$$ }} \)

We'll denote an initial value problem for a differential equation by writing the initial conditions after the equation, as in equation $$(11)$$. For example, we would write an initial value problem for equation $$(2)$$ as

$$y^{(n)}=f(x,y,y', \dots,y^{(n-1)}),\, y(x_0)=k_0,\, y'(x_0)=k_1,\, \dots,\, y^{(n-1)}=k_{n-1}.$$

Consistent with our earlier definition of a solution of the differential equation in equation $$(12)$$, we say that $$y$$ is a solution of the initial value problem equation $$(12)$$ if $$y$$ is $$n$$ times differentiable and

\begin{eqnarray*}
y^{(n)}(x)=f(x,y(x),y'(x), \dots,y^{(n-1)}(x))
\end{eqnarray*}

for all $$x$$ in some open interval $$(a,b)$$ that contains $$x_0$$, and $$y$$ satisfies the initial conditions in equation $$(12)$$. The largest open interval that contains $$x_0$$ on which $$y$$ is defined and satisfies the differential equation is the $$\textcolor{blue}{\mbox{interval of validity}}$$ of $$y$$.

Example $$\PageIndex{6}$$:

In Example $$5$$ we saw that

$$y={x^4+7\over4}$$

is a solution of the initial value problem

\begin{eqnarray*}
\end{eqnarray*}

Since the function in equation $$(13)$$ is defined for all $$x$$, the interval of validity of this solution is $$(-\infty,\infty)$$.

Example $$\PageIndex{7}$$:

In Example $$2$$ we verified that

$$y={x^2\over3}+{1\over x}$$

is a solution of

\begin{eqnarray*}
xy'+y=x^2
\end{eqnarray*}

on $$(0,\infty)$$ and on $$(-\infty,0)$$. By evaluating equation $$(14)$$ at $$x=\pm1$$, you can see that equation $$(14)$$ is a solution of the initial value problems

$$xy'+y=x^2,\quad y(1)={4\over3}$$

and

$$xy'+y=x^2,\quad y(-1)=-{2\over3}.$$

The interval of validity of equation $$(14)$$ as a solution of equation $$(15)$$ is $$(0,\infty)$$, since this is the largest interval that contains $$x_0=1$$ on which equation $$(14)$$ is defined. Similarly, the interval of validity of equation $$(14)$$ a solution of equation $$(16)$$ is $$(-\infty,0)$$, since this is the largest interval that contains $$x_0=-1$$ on which equation $$(14)$$ is defined.

Free Fall Under Constant Gravity

The term $$\textcolor{blue}{\mbox{initial value problem}}$$ originated in problems of motion where the independent variable is $$t$$ (representing elapsed time), and the initial conditions are the position and velocity of an object at the initial (starting) time of an experiment.

Example $$\PageIndex{8}$$:

An object falls under the influence of gravity near Earth's surface, where it can be assumed that the magnitude of the acceleration due to gravity is a constant $$g$$.

(a) Construct a mathematical model for the motion of the object in the form of an initial value problem for a second order differential equation, assuming that the altitude and velocity of the object at time $$t=0$$ are known. Assume that gravity is the only force acting on the object.

(b) Solve the initial value problem derived in part (a) to obtain the altitude as a function of time.

(a) Let $$y(t)$$ be the altitude of the object at time $$t$$. Since the acceleration of the object has constant magnitude $$g$$ and is in the downward (negative) direction, $$y$$ satisfies the second order equation

\begin{eqnarray*}
y''=-g,
\end{eqnarray*}

where the prime now indicates differentiation with respect to $$t$$. If $$y_0$$ and $$v_0$$ denote the altitude and velocity when $$t=0$$, then $$y$$ is a solution of the initial value problem

$$y''=-g,\quad y(0)=y_0,\quad y'(0)=v_0.$$

(b) Integrating equation $$(17)$$ twice yields

\begin{eqnarray*}
y'&=&-gt+c_1, \\
y&=&-{gt^2\over2}+c_1t+c_2.
\end{eqnarray*}

Imposing the initial conditions $$y(0)=y_0$$ and $$y'(0)=v_0$$ in these two equations shows that $$c_1=v_0$$ and $$c_2=y_0$$. Therefore the solution of the initial value problem equation $$(17)$$ is

\begin{eqnarray*}
y=- {gt^2\over2}+v_0t+y_0.
\end{eqnarray*}

Contributors

Trench, William F., "Elementary Differential Equations" (2013). Faculty Authored and Edited Books & CDs. 8.
https://digitalcommons.trinity.edu/mono/8